The document NCERT Solutions (Part- 4)- Algebraic Expressions and Identities Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.

All you need of Class 8 at this link: Class 8

**Exercise 9.5**

**Question 1. **Use a suitable identity to get each of the following products.

**Solution: (i)** Using the identity,

(x + a)(x + b) = x^{2} + (a + b)x + ab

we have

(x + 3)(x + 3) = x^{2} + (3 + 3)x + (3 * 3)

= x^{2} + 6x + 9

**(ii)** Using the identity,

(x + a)(x + b) = x^{2} + (a + b)x + ab,

we have

(2y + 5)(2y + 5) = (2y)^{2} + (5 + 5)2y + (5 * 5)

= 4y^{2} + (10)2y + 25

= 4y^{2} + 20y + 25

**(iii)** Using the identity, (a – b)^{2} = a^{2} – 2ab + b^{2}

(2a – 7)(2a – 7) = (2a – 7)^{2}

= (2a)^{2} – 2(2a)(7) + (7)^{2}

= 4a2 – 28a + 49

**(iv)** Using the Identity, (a – b)^{2} = a^{2} – 2ab + b^{2}, we have

**(v)** Using the identity, (a + b)(a – b) = a^{2} – b^{2},

we have

(1.1m – 0.4)(1.1m – 0.4) = (1.1m)^{2} – (0.4)^{2}

= 1.21m^{2} – 0.16

**(vi)** ∵ (a^{2} + b^{2})(–a^{2} + b^{2}) = (b^{2} + a^{2})(b^{2} – a^{2})

∴ Using the identity, (a + b)(a – b) = a^{2} – b^{2}, we have

(b^{2} + a^{2})(b^{2} – a^{2})= (b^{2})^{2} – (a^{2})^{2} = b^{4} – a^{4}

**(vii)** ∵ (6x – 7)(6x + 7) = (6x + 7)(6x – 7)

∴ Using the identity, (a + b)(a – b) = a^{2} – b^{2},

we have

(6x + 7)(6x – 7) = (6x)^{2} – (7)^{2}

= 36x^{2} – 49

**(viii)** Using the identity, (a + b)^{2} = a^{2} + 2ab + c^{2}, we have

(–a + c)(–a + c) = (–a + c)^{2}

= (–a)2 + 2(–a)(c) + (c)^{2}

= a^{2} + 2(–ac) + c^{2}

= a^{2} – 2ac + c^{2}

**(ix)** Using the identity, (a + b)^{2} = a^{2} + b^{2} + 2ab, we have

**(x) **Using the identity, (a – b)^{2} = a^{2} – 2ab + b^{2} we have

(7a – 9b)(7a – 9b) = (7a – 9b)^{2}

= (7a)^{2} – 2(7a)(9b) + (9b)^{2}

= 49a^{2} – 126ab + 81b^{2}

**Question 2.** Use the identity (x + a)(x + b) = x^{2} + (a + b)x + ab to find the following products.

**(i)** (x + 3)(x + 7) **(ii) **(4x + 5)(4x + 1) **(iii)** (4x – 5)(4x – 1)**(iv)** (4x + 5)(4x – 1) **(v)** (2x + 5y)(2x + 3y)**(vi)** (2a^{2} + 9)(2a^{2} + 5) **(vii) **(xyz – 4)(xyz – 2)

**Solution: **

**Question 3.** Find the following squares by using the identities.

**(i)** (b – 7)^{2 } **(ii) **(xy + 3z)^{2} **(iii) **(6x^{2} – 5y)^{2 }

**(v)** (0.4p – 0.5q)^{2} **(vi) **(2xy + 5y)^{2}

**Solution: **

**(i)** (b – 7)^{2}

Using (a – b)^{2 }= a^{2} – 2ab – b^{2}, we have

(b – 7)^{2} = (b)^{2} – 2(b)(7) + (7)^{2}

= b^{2} – 14b + 49

**Question 4.** Simplify:

**(i)** (a^{2} – b^{2})^{2 }** (ii) **(2x + 5)^{2 }– (2x – 5)^{2}

**(iii)** (7m – 8n)2 + (7m + 8n)^{2} **(iv)** (4m + 5n)^{2} + (5m + 4n)^{2}

**(v)** (2.5p – 1.5q)^{2} – (1.5p – 2.5q)^{2} **(vi)** (ab + bc)^{2} – 2ab^{2}c

**(vii) **(m^{2} – n^{2}m)^{2} + 2m^{3}n^{2}

**Solution:**

(i) (a^{2} – b^{2})^{2} = (a^{2})^{2} – 2a^{2} * b^{2} + (b^{2})^{2} [using (a – b)^{2} = a^{2} – 2ab + b^{2}]

= a^{4} – 2a^{2}b^{2} + b^{4}

**Question 5.** Show that

**Solution:**

**(i) **LHS = (3x + 7)^{2} – 84x

= (3x)^{2} + 2(3x)(7) + (7)^{2} – 84x

= 9x^{2} + 42x + 49 – 84x

= 9x^{2} + (42 – 84)x + 49

= 9x^{2} – 42x + 49

RHS = (3x – 7)^{2}

= (3x)^{2} – 2(3x)(7) + (7)^{2}

= 9x^{2} – 42x + 49 = LHS

Since, LHS = RHS

∴ (3x + 7)^{2} – 84x = (3x – 7)^{2}

**(ii) **LHS = (9p – 5q)^{2 }+ 180pq

= (9p)^{2} – 2(9p)(5q) + (5q)^{2} + 180pq

= 81p^{2} – 90pq + 25q^{2} + 180pq

= 81p^{2 }+ (–90 + 180)pq + 25q^{2}

= 81p^{2 }+ 90pq + 25q^{2}

RHS = (9p + 5q)^{2}

= (9p)^{2} + 2(9p)(5q) + (5q)^{2}

= 81p^{2} + 90pq + 25q^{2}

Since, LHS = RHS

∴ (9p – 5q)^{2} + 180pq = (9p + 5q)^{2}

**Question 6.** Using identities, evaluate:

(i) 71^{2} (ii) 99^{2} (iii) 102^{2} (iv) 998^{2} (v) 5.2^{2} (vi) 297 * 303 (vii) 78 * 82 (viii) 8.9^{2} (ix) 1.05 * 9.5

**Solution: **

**(i) **∵ (71) = (70 + 1)

∴ (71)2 = (70 + 1)^{2}

= (70)^{2} + 2(70)(1) + (1)^{2} [Using (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 4900 + 140 + 1 = 5041

**(ii) **We have 100 – 1 = 99

∴ (99)^{2} = (100 – 1)^{2}

= (100)^{2} – 2(100)(1) + (1)^{2} [Using (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 10000 – 200 + 1 = 9801

**(iii) **∵ 102 = 100 + 2

∴ (102)^{2} = (100 + 2)^{2}

= (100)^{2} + 2(100)(2) + (2)^{2} [Using (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 10000 + 400 + 4 = 10404

**(iv)** ∵ 998 = 1000 – 2

∴ (998)^{2} = (1000 – 2)^{2}

= (1000)^{2} – 2(1000)(2) + (2)^{2} [Using (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 1000000 – 4000 + 4 = 996004

**(v)** ∵ 5.2 = 5 + 0.2

∴ (5.2)^{2} = (5 + 0.2)^{2}

= (5)^{2} + 2(5)(0.2) + (0.2)^{2} [Using (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 25 + 2 + 0.04 = 27.04

**(vi)** ∵ 297 = 300 – 3 and 303 = 300 + 3

∴ 297 * 303 = (300 – 3)(300 + 3)

= (300)^{2} – (3)^{2} [Using (a + b)(a – b) = a^{2} – b2]

= 90000 – 9 = 89991

**(vii) **∵ 78 = 80 – 2 and 82 = 80 + 2

∴ 78 * 82 = (80 – 2)(80 + 2)

= 80^{2} – (2)^{2} [Using (a + b)(a – b) = a^{2} – b^{2}]

= 6400 – 4 = 6396

**(viii)** ∵ 8.9 = (9 – 0.1)

∴ (8.9)^{2} = (9 – 0.1)

= (9)^{2} – 2(9)(0.1) + (0.1)^{2} [Using (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 81 – 1.8 + 0.01

= 81.01 – 1.80 = 79.21

**(ix) **∵ 1.05 = 1 + 0.05

∴ (1.05) * 9.5 = (1 + 0. 05) * 9.5

= (1 * 9.5) + (9.5 * 0.05)

= 9.500 + 0.475 = 9.975

**Question 7.** Using a^{2} – b^{2} = (a + b)(a – b), find:

(i) 51^{2} – 49^{2} (ii) (1.02)^{2 }– (0.98)^{2} (iii) 153^{2} – 147^{2} (iv) 12.1^{2} – 7.9^{2}

**Solution: **

**(i) **51^{2} – 49^{2} = (51 + 49)(51 – 49)

= (100)* (2) = 200

**(ii) **(1.02)^{2} – (0.98)^{2} = (1.02 + 0.98)(1.02 – 0.98)

= (2.0) * (0.04) = 0.08

**(iii)** 153^{2} – 147^{2} = (153 + 147)(153 – 147)

= (300) * (6) = 1800

**(iv) **(12.1)^{2} – (7.9)^{2} = (12.1 + 7.9)(12.1 – 7.9)

= 20 * 4.2 = 84

**Question 8.** Using (x + a)(x + b) = x^{2} + (a + b)x + ab,

find: (i) 103 * 104 (ii) 5.1 * 5.2

(iii) 103 * 98 (iv) 9.7 * 9.8

217 videos|147 docs|48 tests

### Subtraction of Algebraic Expressions

- Video | 04:37 min
### Extra Questions- Comparing Quantities

- Doc | 1 pages
### Examples: Algebraic Expressions(Addition, subtraction)

- Video | 07:01 min
### Introduction to Multiplication of Algebraic Expressions

- Video | 06:07 min
### Multiplying 2 Monomials

- Video | 08:23 min
### Multiplying 3 or more Monomials

- Video | 05:33 min

- Addition of Algebraic Expressions
- Video | 03:17 min